Ex 12.2, 5 - In a circle of radius 21 cm, an arc subtends 60 - Area of segment of circle and length of arc

  1. Chapter 12 Class 10 Areas related to Circles
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Ex 12.2, 5 In a circle of radius 21 cm, an arc subtends an angle of 60ยฐ at the centre. Find: the length of the arc Length of Arc APB = ๐œƒ/360ร—(2๐œ‹๐‘Ÿ) = (60ยฐ)/(360ยฐ)ร—2ร—22/7ร—21 = 1/6ร—2ร—22/7ร—21 = 22 cm Ex 12.2, 5 In a circle of radius 21 cm, an arc subtends an angle of 60ยฐ at the centre. Find: (ii) area of the sector formed by the arc Area of sector OAPB = ๐œƒ/360ร—๐œ‹๐‘Ÿ2 = 60/360ร—22/7ร—21ร—21 = 1/6ร—22/7ร—21ร—21 = 1/6ร—22ร—3ร—21 = 231 cm2 Ex 12.2, 5 In a circle of radius 21 cm, an arc subtends an angle of 60ยฐ at the centre. Find: (iii) area of segment formed by the corresponding chord Area of segment APB = Area of sector OAPB โ€“ Area of ฮ”OAB From last part, Area of sector OAPB = 231 cm2 Finding area of ฮ” AOB Area ฮ” AOB = 1/2 ร— Base ร— Height We draw OM โŠฅ AB โˆด โˆ  OMB = โˆ  OMA = 90ยฐ In ฮ” OMA & ฮ” OMB โˆ  OMA = โˆ  OMB OA = OB OM = OM โˆด ฮ” OMA โ‰… ฮ” OMB โ‡’ โˆ  AOM = โˆ  BOM โˆด โˆ  AOM = โˆ  BOM = 1/2 โˆ  BOA โ‡’ โˆ  AOM = โˆ  BOM = 1/2 ร— 60ยฐ = 30ยฐ Also, since ฮ” OMB โ‰… ฮ” OMA โˆด BM = AM โ‡’ BM = AM = 1/2 AB From (1) AM = 1/2AB 2AM = AB AB = 2AM Putting value of AM AB = 2 ร— 21/2 AB = 21 Now, Area of ฮ” AOB = 1/2 ร— Base ร— Height = 1/2 ร— AB ร— OM = 1/2 ร— 21 ร— โˆš3/2 ร— 21 = (441โˆš3)/4 cm2 Area of segment APB = Area of sector OAPB โ€“ Area of ฮ”OAB = (231 โ€“ 441/4 โˆš3) cm2

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